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How do I intentionally fly into turbulence?
How could I plan a flight to increase my chances of a fun ride through turbulence?  What if I want to plan my flight before the weather forecast is available?  The internet is full of people who are afraid of turbulence, but I find chop enjoyable, within reason.
Edit:  I am still looking for answers like

Plan your flight to a location with [geographic properties] at [altitude] feet when it is [time of day] in the [season] and you are likely to find a reasonable amount of turbulence.

I think these are wrong answers:  Fly into a thunderstorm cell.  Violate minimum separation with an A380.

Maybe there are some mountains that always generate turbulence?

The problem with turbulence is that you can't be sure of the "within reason" part -- especially in a light plane, it can go from "a little bumpy" to "why am I upside down with the wing stalled and how did I lose 2000 feet of altitude in no time?" without warning...

With plan a flight, do you mean as a passenger on a scheduled flight or as a private pilot picking a route?

The difference between your idea of fun and fatal in a light airplane could be very small. Be careful what you wish for.

@Sanchises Passenger's flight planning would be off topic here, right?

@GdD Aircraft type could be part of a good answer.

If you like turbulence you should consider aerobatics, it's way better.

On the extreme side, join the Air Force, specifically 53rd Weather Reconnaissance Squadron. It is their job to fly into the center of hurricane. The NOAA Hurricane Hunters is a civilian agency that do similar research.

The problem seems to be right there in your wording: "the chances".

@GdD and Zeiss Ikon, the OP already said that known dangerous situations like thunderstorms and wake turbulence are not good answers.  Airplanes fly through "fun" turbulence every day but we don't read about planes crashing all the time because of it. Sure there are known incidents, just like people have been injured or killed on roller coaster rides. But I don't think the OP would be risking their life by purposely flying into an area that has expected turbulence.

@StevePemberton, you are probably right, however we've had people ask how to do something that is reckless and dangerous to themselves or others on this site more than once. I am not going to make the assumption that someone knows what they are really asking for.

@GdD - In the question the OP excluded weather so they are not intending to fly into dangerous weather conditions. They are looking for geographic locations where turbulence is more likely. The "within reason" qualifier I think makes it clear that they are not looking for answers that would be reckless or dangerous. Maybe they could have been more explicit about that, but I'm pretty sure that they are talking about the type of turbulence that is normal and safe, but just not what most people find to be desirable.

The best/safest way I know to get "within reason" turbulence is to go flying on a day when the sailplanes are having good flights.  Every little popcorn cumulus cloud will have a thermal under it, and that thermal will give you one or two nice bumps as you fly through it, but they'll be gentle enough (providing your avoid budding cumulonimbus) not to damage you or even a pretty lightweight airplane like a Cub or Pacer.  Do keep a very careful eye out for unpowered traffic in that weather; they have right of way and can be hard to spot, since they're likely to be above or below you, not just off to the side or ahead.
The problem is that the weather that creates those cloud lanes and popcorn in the sky can also change to a thunderstorm (with "beyond reason" turbulence, updrafts that can suck you to 20,000 feet or higher while you try to keep the wings from coming off, and lightning that can flip every breaker on your panel, hail that can punch right through a light plane, etc.), so in those conditions you need to keep an eye on the weather and an ear on weather reports from folks with better instruments than you.
Where you'll get turbulence related to what glider pilots call "lift" is two classes of areas: mountain ranges perpendicular to current wind direction, creating ridge lift (which will be uneven due to the uneven nature of the mountain front) and large flat regions in sunlight, especially if there is a patchwork of plowed and grown-over fields (plowed earth will warm faster and get warmer than crops, producing more thermal activity).

Good advice.  However, in my experience most places do not have any sailplanes.  Wikipedia's old information says there are only 33,000 gliders.

@AnonymousPhysicist While this is true, it doesn't change the kind of weather glider pilots like -- they need rising air to stay up, and if you don't have hills or mountains to provide ridge lift, you need thermals.

@AnonymousPhysicist That's not a correct understanding of statistics. Areas that are known to be good weather for sailplanes will naturally attract them, so those areas will have much higher expected concentration of them than anywhere else.

and that thermal will give you one or two nice bumps as you fly through it. On good soaring days thermals can give much much more than "one or two nice bumps", and even more if you are closer to the ground on a windy day.

@Nelson The answer is about when there is turbulence, not where.  Gliders relocate very slowly, so I think my comment is accurate.  I will edit the question to encourage answers about "where."

What is your definition of turbulence? If it's crazy enough, it could be dangerous it fly into. That's the turbulence we mostly know where it is (because flight wouldn't be safe without that knowledge). That includes thunderstorms and wakes from large aircraft.
There is a good band of turbulence that we can't predict, but is typically not dangerous.
There is some like Zeiss Ikon mentioned that is usually OK and somewhat predictable (updrafts that condense at the top, thus being marked with a cloud).
Then there is some that is pretty light and predictable: your own wake. Fly a steep turn circle, and if you do it right, you'll hit a bump that is your own wake.
And, yes, mountains + wind does make turbulence. Just like water flowing over a rock in a river, you get a repeating wave downstream of the mountain/rock. It sweeps into a peak and then drops again, sweeps up into a peak, etc. Near the top of these, there can be rotors, sometimes even rotor clouds, and that is the worst turbulence I have experienced in a GA aircraft: we were in glass smooth updraft and then all of the sudden we rolled until the CFI thought we were going to go into an upset attitude.
Is your definition of turbulence the wings rolling a bit and the nose bumping up and down, or more of a "your stomach while on a roller coaster" type?
If it's the roller coaster feeling, you can make that feeling with how the aircraft is flown. Go a ways up and fly something not quite parabolic arc for something like the swing feeling, or a perfect parabolic arc for a weightless feeling (yes, I can tell you the two a different, from skydiving experience).
If it's the random rolling/pitching, you could just pay for a flight lesson in a GA aircraft, and (at a good margin less than Va) ask the CFI to do their best to simulate (while staying safe) the worst turbulence they've ever experienced, with random inputs. If that's not good enough, do something similar with an aerobatic pilot in an aerobatic plane. If that's not good enough, see if you can somehow arrange a ride with a fighter pilot with a similar request. I highly suspect that somewhere along that progression, you'll find that "that's enough" point!

Dangerous turbulence is excluded with the OP's qualifier "within reason". Turbulence by its nature involves unpredictable movement in all directions, which is part of the appeal to those who find it fun (at least it is for me). Parabolic arcs, or aerobatics as was also suggested would not be the same thing. My definition of fun is a level where for example it is difficult or even impossible to walk down the aisle of an airliner because of pitching in all directions. Although I only find that fun if the flight attendants and all passengers are safely seated. Mountains seems to be a good answer.

@StevePemberton I think one of the things that makes answering this question hard is we are lacking some context: is this person a pilot or just a part 121 (scheduled airline) passenger? "fun ride" sounds like 121 passenger, but "plan my flight" sounds like part 91 pilot.  The "common" sense needed to evaluate "within reason" could very well not be in common between those groups. Remember that freight pilots say "boxes don't bitch" where part 121 pilots work to to avoid (escape when encountered) turbulence freight would consider minor.

Azendale - I agree on first reading it's a little vague, but the needed information is there. The OP pointed out in a later comment that passenger travel would be off-topic, and in fact in the question they used the term "plan a flight" instead of select a flight. Passengers plan trips, pilots plan flights. And the lack of specificity about the level of turbulence other than the "within reason" qualifier means that while dangerous extremes are excluded, the OP is not limiting answers to a specific level of turbulence. And the "fun" qualifier implies more than just making your coffee swirl.

@StevePemberton Some of this question could be "how can I replicate a feeling I haven't experienced any other way". Both freight and airlines are large planes less affected by turbulence, than a lighter, small, low wing loading GA aircraft, like a Kitfox 7. So has OP experienced GA turbulence? Note that extremes is not in one scale: you have extremes of safety and extremes of how it feels. I have been in turbulence that didn't feel very weird at the time, but looking back it it was actually dangerous. Is having a buddy input large random inputs in a GA plane well under Va a valid answer?

Azendale - since they didn't specify a level of turbulence I don't think they are limiting it to levels that they have personally experienced in whatever type of flying they have done. I interpret "within reason" as relating to safety not the experience. But that's just my interpretation. Your didn't feel weird example sounds like the turbulence itself wasn't dangerous, it was the weather conditions that you were in that were dangerous which you realized later. The question excludes dangerous conditions, as clarified by the OP in a later comment.

@StevePemberton You are correct on all counts.  I don't know why so many people didn't understand the question.

You can create your own turbulence and fly through it, although it will be very short in duration.  As a Navy T-6 instructor pilot I had this happen routinely on aerobatic flights.  When a proficient student would perform a loop or a steep 360 degree turn correctly the result was us flying through either our own prop wash, or wing tip vortices.  The result of this is a moderate bump when you hit it.  Not the same as sustained turbulence, but still pretty cool.

Nice practical answer.


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What is the airspeed record for a helicopter flying backwards?
How fast can a rotocraft go, relative to the air, if it is flying towards its tail?
Tiltrotor and powered lift aircraft do not count.  Helicopters and gyroplanes do.  The aircraft must have a readily visible tail end, so the  Jerome-de Bothezat Flying Octopus for example is excluded.  This is about level flight with zero yaw.

https://aviation.stackexchange.com/questions/15853/what-is-the-fastest-recorded-backward-flying-plane

Given that a rotorcraft will tend to weathercock into wind I suspect that speed is more due to the skill of the pilot than the capability of the aircraft. The FAI don't appear to have a class for 'Helicopters flying backwards'. Perhaps you should set one.

@CatchAsCatchCan I agree skill is important. The ideal answer would identify the particular flight and pilot, not just the type.  I think flights with an autopilot should count, but I don't see why anyone would develop an autopilot that could set the record.

Would a Chinook be indifferent about flying cockpit-forward or ramp-forward?

@CatchAsCatchCan --re "Given that a rotorcraft will tend to weathercock into wind" -- can't be.  That is all pilot-induced.  (Unless maybe you meant the "relative wind".)

@MichaelHall Flying forward and then yawing 180 degrees does not count as flying backwards.

African or European?

@JonCuster You have to know these things when you're a king.

Inspired by Jpe61's answer I checked the flight manual of the Chinook, and it mentions
"Maximum airspeed in rearward flight is 45 knots" (23.15 m/s)
This value is rather close to the upper bound of Jpe61's estimate.

Of course, this does not mean it's the maximum speed achieved, but I think the Chinook is a helicopter whose setup is quite well suited to flying backward, and this will be close to the maximum speed achieved.

What? A helicopter can go backwards faster than 10 knots or 5 m/s

I thought so too, but I can't find any measured numbers/sources confirming it, so I'm putting this as a lower bound.

Since the video i linked to is from an airshow, it makes sense they are not using the full envelope to have safety margins

This video:
Chinook CH-47 Roll and flying backwards amazing to see at RIAT 2017 shows a Chinook taxiing backwards and the proceeding to take off at the same backwards speed. I estimated it travelled its length in about 1.5 to 2 seconds, and knowing the length being 30 meters, that would give it a speed of approximately 15 to 20m/s, which would be about 30 to 39kts.
Chinook is, of course, not your garden variety helicopter, its tail end is clearly distinctive though.

chinook is also the only helicopter where flying backwards makes sense, turning around it's vertical axis is trivial for other helicopters.

For the maximum reverse speed of a conventional helicopter I have found one example, the Bell AH-1F Cobra.
The "Technical manual operators manual for army model AH-1F attack helicopter" sets the rearward flight speed limit as 30 kts.
This is lower than the sideways limit which is 35 knots, and I agree with Sophit in that the tail becomes destabilizing factor in rearward flight, restricting the envelope.
(Section V. airspeed limits, page 163)

I don't have hard numbers (although the maximum backward speed should be readily available in the flight manual of any helicopter) but if you consider helicopters with tails then some 20 kt should be a maximum limit. A tailplane is added to give the aircraft stability and download the rotors... when flying forward! Flying backward the tailplane becomes destabilising and the rotors now need to provide additional stability as well, more and more the faster they fly backwards.

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Tail Warning Radar on the P-51
One of the "features" of the P-51 was a tail warning radar that was supposed to let pilots know if an enemy fighter was on your tail.
However, a P-51 pilot told me that they generally disabled them.
According to him, the problem was that, if you were flying in formation and performing a turn, there was generally a point when a squadron-mate would be behind you. When that happened, a loud horn would sound and a bright light would flash in your eyes, scaring the bejesus out of you.
Has anyone else heard of that problem?
EDIT
You are right, it was probably a bell - I heard the story 35 years ago. And the fact it had volume controls indicate that he would have turned the TWR down/off rather than disconnecting it.
I have read mixed reviews regarding the gunsight. Some of the favorable reviews were from old-timers who I thought would hate it.
For those who doubt that the P-51D had TWR, here is a picture from the POH showing the controls for the TWR (aka Rear Warning Radar), which are referenced in the comments below - item 8.

The AN/APS-13 had a volume knob and an intensity knob for the light, as well as an on/off switch. It was likely only turned on when the possibility of combat was present, but I can't find any anecdotal evidence of that or this claim.

I recall that it was a bell not a horn, but maybe switched to a horn later?  Pilots didn't like the K-14 lead computing gunsight either and would only use the fixed backup/ground attack reticle, except for some who took the trouble to master use of the gyro driven reticle.  It allowed you to delete your tracers from the loadout so your target was unaware you were shooting at them until you were scoring hits, a huge tactical advantage. Gun camera film with no tracer apparent means the pilot is aiming with his K-14's gyro reticle.

@RalphJ Tail warning radar was widely available in 1944.   just watch a DCS sim video on youtube.  The DCS P-51 has it.  The red light on the left side of the glareshield.

...and here I thought they just had a trained monkey who would pull the string for the bell if he saw anything coming up behind you. LOL!  I had no idea P-51's had this.

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P-51 Supercharger Loops
A WWII pilot once told me that, for fun, they would sometimes take their P-51s up and perform what he called "supercharger loops". They would climb to an altitude below where the supercharger would switch from low to high and then start performing loops. The idea was that they would all adjust their superchargers so that they all switch from low to high and back at the same time. And, eventually, they would be flying synchronized loops.
Has anyone else heard of this?
Is it possible to adjust the changeover speeds from inside the cockpit? Or did that have to be done in the shop?
(There is a supercharger switch in the cockpit, but that merely appears to be a toggle switch that you could use to over-ride the automatic settings.)


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Which Provides a Faster Turn - A Horizontal Turn or a Vertical Loop?
I am starting to research this question with respect to WWII aircraft - specifically the FM2 Wildcat, which was considered an excellent dogfighter.
To approximate an "apples to apples" comparison:

For horizontal turns, I am using the maximum sustained turn rate that results in no net altitude loss. The aircraft will remain at this speed and altitude and turn rate for the entire time.
For vertical loops, I computed the rate of turn for each speed that will return the aircraft to the same speed and altitude at the bottom of the loop.  Since speed and rate of turn (both horizontal and vertical) vary throughout the loop, I am using the average rate of turn for the entire loop. For sake of simplicity, the loop is performed using the same angle of attack for the entire loop.

I created the above E-M Chart for the aircraft. As is typical, the performance envelope is bounded by the minimum speeds for the aircraft and the maximum g-load - which I assume to be around 7g's. In level sustained flight, the g-limit does not ever come into play because the horsepower of the aircraft is simply insufficient to push the aircraft around in high-g maneuvers. The maximum g's in a sustained turn are around 3 g's.
Within the chart, I plotted the sustained turn rate for level flight at various speeds (PS=0).  As shown, as speed increases, the turn rate initially increases (along the Vmin line) and then decreases. At top speed, Thrust = Drag and Lift = Weight.  As speed increases, less angle of attack is required to produce the Lift, which means that induced Drag also decreases. However, parasitic drag is also increasing - to the point that Parasitic Drag = Thrust - Induced Drag.  At this speed, all of the Lift is used to keep the aircraft in level flight.  There is no excess Lift which can be used in level turning flight. Any effort to turn the aircraft while maintaining altitude will result in the aircraft slowing down to the speed at which the desired rate of turn and altitude can be sustained.
The same kind of thing happens in a loop. When you increase the angle of attack to the desired amount, this creates more Lift and also more Induced Drag.  This Induced Drag along with the fact that the aircraft is climbing and being slowed by gravity - will cause the aircraft to slow down. The aircraft will reach the lowest speed and fastest turn rate at the top of the loop. But, in contrast with a horizontal turn, this lost speed can be recovered in the last half of the loop, where the aircraft is descending.
As with the horizontal turn, the average turn rate for a loop decreases with speed. However, unlike a horizontal turn, the average turn rate never goes to zero. You can still perform a loop even when the aircraft is at maximum speed. The aircraft will lose speed during the first half of the loop and return to maximum speed at the bottom of the loop.
But, what was more surprising is that the vertical loop always yields a better average turn rate than the horizontal loop. This indicates that, if you are engaged in a turning contest with another aircraft, you should use vertical loops through the entire range at which they are available.
I have plotted the turn rates for loops on the E-M Chart.  Note that the results appear linear.
QUESTION
My question is whether this result makes sense from a mathematical and real world sense.
I assume that others have made similar computations for other airplanes. The math uses the basic force equations for Thrust, Parasitic Drag, Induced Drag and Lift. I converted these forces to acceleration by dividing force by mass. I netted the acceleration from Thrust against Drag and the portion of Gravity working along the direction of flight (gravity X sin(pitch)). I netted Lift and the portion of gravity opposing Lift (gravity X cos(pitch)) and treated the net Lift as a flight path "deflector" where the angle of deflection = Net Lift X 360/(2*PI()*Speed). This last equation treats the Net Lift as operating along a circular path.
The turn rates for horizontal flight could be easily computed for each different speed. But because the speeds and turn rates for loops are constantly changing, I used a series of computations in 1/60 second intervals (on an excel spreadsheet).  I tried different Angles of Attack until I found the one that resulted in the same speed for the beginning and end of each loop.
In real life, I have flown simulated air combat in Marchetti aircraft with F-16 safety pilots. We started using horizontal turns and quickly switched to using loops. We had been practicing 3-4 g horizontal turns. It seemed that the loops involved about the same g-load (except for the one case where I decided to extend the downward part of a loop before pulling up - that was 5.5 g's).
I also tend to use loops in computer air combat simulations.
Another factor that seems to tip the advantage in favor of the loop is that you have the advantage of being able to use bank to quickly change heading, especially when you are pointed straight up or straight down.
ADDITION
One of the comments below indicates that it would be helpful to provide a summary of my results for a particular speed, e.g. 250 mph:
deg     frame   dif     mph     cL
000.00                  250.00  0.8181
090.00  309.8   309.8   168.08  0.8181
180.00  592.5   282.8   124.87  0.8181
270.00  852.3   259.8   201.65  0.8181
000.00  1127.7  275.5   250.00  0.8181
Recap:
total time          18.796 secs
average turn rate   19.153 deg/sec
average speed       187.44 mph
diameter of turn    1709 ft

This shows the speeds at various points on the loop and the time needed to reach each point (1 frame = 1/60 sec).
Regarding g-forces, my computations show that the cL for level flight at 250 mph is around 0.1696 (an Angle of Attack of around 1.696 deg). Thus, a cL of 0.8181 (an Angle of Attack of around 8.181 deg) should result in about 5 g's of force. At the top of the loop, the aircraft speed is 124.87 mph, well above stall. In real life, the pilot will likely vary the Angle of Attack, as suggested in the comments. However, it appears that (at least on the Fm2) a constant Angle of Attack does not break any limitations and should yield an approximation of the turn rate - possibly on the low side.
FM2 vs. ZERO (edited Jan 26, 2023)
The question of who would win a dogfight in a contest between an FM2 and Zero is interesting. For those who are interested there was an official comparison of these aircraft during WWII. The comparison concluded that: "The Zeke 52 could gain one turn in eight at 10,000 feet." and that the FM2 was more maneuverable at speeds above 200 kts (230 mph). The report recommended "DO NOT DOG-FIGHT WITH THE ZEKE 52." They made the same recommendation for the Hellcat.

I created an E-M Chart for the A6M5 Zero Model 52, a mid-war improved Zero. The FM2 and A6M5 are remarkably similar. Both have about the same top speed (around 300 mph) in level flight, about the same power to weight ratio and about the same wing area - although the A6M5 is lighter and, therefore, has a lower wing loading.
A comparison of the horizontal turn rates shows that, if both fighters are turning at the same G-loads, the A6M5 has a very slight advantage in turn rate. However, the A6M5 has a significant advantage if both are attempting to turn and maintain the same altitude. If you compare the charts, you will see that the level turn rate (PS=0) for the A6M5 is above the 3g line. The exact calculations are that, in a full circle, the A6M5 will gain 50.53 degrees, or one turn in seven, at sea level. The lower wing-loading seems to be the key factor in this difference.
But a comparison of vertical turn rates shows an interesting result. In this case, the FM2 now has a small advantage over the A6M5. Since lift plays less of a role in a vertical turn, this appears to be a result of the slightly higher power to weight ratio of the FM2.
The FM2 was a higher performance version of the F4F Wildcat. Thus, as with the Hellcat, unsuspecting Zero pilots might have been surprised by this improved vertical performance, to their detriment. This is reflected in the 32:1 K/D ratio of the FM2.
TENTATIVE CONCLUSION
This has been an interesting discussion, although not what I expected when I asked the questions.  Nevertheless, I have learned enough to tentatively conclude that my two questions have been answered.
First, with regard to real life experiences, it appears that vertical loops are not used tactically because of various limitations and complications. The limitations are that utilizing them requires that you complete at least 1/2 of the loop. The complications are that you now have to figure out how to re-engage with the enemy. It could be that vertical loops make more sense with smaller prop-driven aircraft, like the Marchetti or FM2. In the case of the Marchetti, we were in similar aircraft and the other person started doing vertical loops first - which meant we had to follow. In the case of the FM2, it has been pointed out that this would have been a losing strategy against the Zero.
Second, regarding the math, I have seen nothing to indicate that my computations are wrong - only that other factors outweigh any possible turn rate benefits. Vertical half-loops have been used to quickly change direction and escape from combat.  Two examples are the Immelmann, a climbing vertical half-loop first used by the Germans in WWI and the Split-S, diving vertical half-loop first used in WWII and still used today.
So I will mark this question as answered.
Note:
Since writing this conclusion, an additional response has been received which focuses more on the math and is also very helpful pointing out that the reason the turn rate for the loop is faster is because all of the lift is directed towards creating the turn rate - in contrast with a horizontal turn, where some of the lift is used to offset gravity.
I have also replaced the E-M diagram with a diagram that uses the recognized term PS=0 to refer to a horizontal turn where vertical speed is zero. Also the description for the vertical axis has been modified to clarify that the E-M diagram measures the horizontal turn rate.

If an FM2 got into a turning dogfight with a Zero, it was toast. A Zero could escape a WC on its tail by going straight up into a hammerhead turn, knowing the WC would run out of energy first if the WC followed, and the Zero would complete its hammerhead perfectly positioned on the WC's tail below it.  It worked like a charm time after time until one day a famous Zero ace did the maneuver, but unbeknownst to him the airplane following him up was a Hellcat, and that was the end of that Zero. It was the first Japanese Hellcat encounter. All downhill from there.

Keeping AoA constant throughout a loop will not optimize turn performance.  You will either over-G initially, or not pull enough over the top while you are slower.  (perhaps depending on the airplane...)

Very similar in dimensions and top speed.  Zero 1000 pounds lighter, not as much power, slightly less wing area.  Looking at generic POH, pulling turns are limited by stalling and excessive G's.  Given equal pilot tolerance to Gs and strong aircraft, the slower, lighter aircraft turns faster.  Interestingly, chopping throttle, the lighter plane slows down faster and has greater acceleration into the turn.  The heavier plane stalls easier.  Amazingly, biplanes more than held their own in turning as compared to the newer planes.  The monoplane pilots were told to "stay away from them".

@RobertDiGiovanni I've read various Fairey Swordfish accounts that claim that thair ability to "flit to and fro" (advanced technical terminology) made them able to evade much more capable monoplane competition.

@RussellMcMahon yea, and a super high wing loaded supersonic jet wouldn't exactly be good at that.  But that's why tactics evolved.

It could be that Zero pilots loved doing vertical loops - possibly because they yielded a better turn rate, or possibly because they could use them to lure Allied pilots into a turning contest which the Allied pilot could not win (no more than they could have won a horizontal turning fight against a Zero).

I have edited the discussion of the FM2 vs. Zero to add an E-M chart for the Zero and to explain the results of a comparison of the two charts. As expected, the Zero has a significant advantage in a horizontal fight. However, in a vertical fight, the FM2 has a slight advantage, similar to that enjoyed by the Hellcat.

Bottom line: you need to keep the aircraft as close as possible to its "Corner Velocity", which is the highest point on its energy maneuverability diagram. (Note, all this comes from an innovative USAF fighter pilot named John Boyd, who more or less invented EM Theory). It is at this point (Corner Velocity), where the turn rate is at its maximum, and the turn radius is at its minimum. Because this point is almost guaranteed to result in a very high negative specific Energy that means that the aircraft will lose energy at that point, but instantaneously, it will be at the highest turn rate and smallest turn radius possible. So, to turn 180 degrees in the fastest amount of time, you want to keep it there as long as possible. That means that you have to turn in a 2D plane that loses a massive amount of energy but slows down the least. The way to do that is by losing energy in the form of altitude instead of airspeed. i.e., you need to descend as you turn.
In the F-4 we practiced this maneuver. Unlike most modern aircraft (F-15, F-16, etc.), that can sustain 7 Gs in a level turn, the F-4 could only sustain about 4-5 Gs in level flight and would lose energy rapidly at high turn rates and tight turns. So, to practice turning around as fast as possible, we would start at or slightly above corner velocity (420 KCAS in the Hard-wing F-4, 380 KCAS in the LES F-4), engage full afterburner, roll the aircraft into 135 degrees of bank, and pull on the pole (the flight control stick) to the maximum allowable AOA, and hold it while turning the aircraft through 180 degrees heading change in a 45 degree inclined plane. In performing this maneuver, we would lose about 4-5000 feet of altitude, but would maintain at or near the original entry airspeed. it would take about 12-14 seconds. Modern aircraft (F-15s, F-16s, etc.) can perform this way faster.
For each more advanced aircraft (F-15, F-16, F-22, etc.) we would need some veteran who flew that aircraft to tell us exactly what flight path angle (plane of turn) would allow him/her to perform a continuous 180 degree turn in that plane without losing airspeed. I cannot say with any certainty, but my suspicion is that even the most advanced aircraft are not capable of doing that without slowing, or descending a bit... Performance advances in modern aircraft have not only increased their thrust to weight ratios and turn rates, they also have corresponding increases in maximum capable AOA and resultant drag that cause enormous energy bleed rates at high turn rates.

This is a good conversation, but comments are not for extended discussion; this conversation has been moved to chat.

Short answer, the vertical loop will provide a faster rate of turn.
For the moment, "stall turns", valuable as they are, will be set aside so that basic fundamentals can be discussed.

a turning object is constantly accelerating laterally.

Increasing force into the turn increases rate of turn.  Decreasing weight increases rate of turn.  What?  That's not what the rate of turn formula says:
ROT = tangent bank angle/Velocity
This is only true when the heavier aircraft can increase its AoA.

at maximum AoA and a given speed (therefor a given amount of lift), the heavier object has a lower lateral acceleration.

Can we just go faster?

That increases the radius of the turn proportionally to the rate of lift increase (both based on V$^2$).

Can we turn faster by increasing Coefficient of Lift

Absolutely!  This increases lift at a given speed, allowing higher bank angle in a horizontal turn.  Dropping slats and flaps increases coefficient of lift.  A real eye popper comes when an F-14 Tomcat extends its wings, enabling it to turn inside even an F-16.

isn't it smarter to use the entire lifting force of the wing into the turn?

This is the advantage of the loop.

in a horizontal turn, some of the lift must be used to maintain altitude.

In a loop all of the lifting force is into the turn, and, at the top of the loop, gravitational force is also into the turn.  Hence the "egg" shaped path through the sky.

Does the "split S" work the same way?

Yes.  Essentially, a half roll puts one "at the top of the loop", ready to use gravity to help out.  Airspeed and G forces may become excessive, but the initial rate of turn (with gravity) is rapid.

Is rate of turn really lower at slower speeds?

Only if some of it has to be apportioned for a constant amount of lift.  One other thing to note is that planes of that era were propeller driven, meaning more thrust at lower airspeeds and very strong engine torque turning tendencies that could be put to use.
So, why not always loop?  Slowing down and turning can be dangerous if a second or third plane is coming.  This is why there is no disagreement here about the "corner speed" concept.  Corner speed shows up in a Cessna 172 POH as Va, manuevering speed.

This is a good conversation, but comments are not for extended discussion; this conversation has been moved to chat.


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Did LeMay actually require crews to remove their guns on B-29s?
There is a persistent rumour/storyline that, during and after the successful low-level raids on Tokyo in March 1945, LeMay ordered that all B-29s remove their guns, except for the tail gun. However, I have found no evidence that this ever really happened.
Removing gun turrets would have left holes in the fuselage. And removing guns would have left a hole in the gun turrets. I have been told that they might have been encouraged to limit their supply of ammo (to reduce weight). However, everyone appears to have ignored this order. Similarly, as far as I can tell, the crews never gave up their gunners. (Interestingly, though, gunners were sometimes called "observers", implying that their primary responsibility was now to observe the condition of the aircraft from their stations.)
I am intimately familiar with the operations of the 6th Bomb Group stationed on Tinian and have talked to many veterans of this and other Groups. However, I am not as familiar with operations on Guam, where LeMay was headquartered. Does anyone know whether the Groups on Guam or elsewhere were required to remove their guns?
ADDENDUM (REVISED)
Regarding several of the comments - you are correct in surmising that the guns and/or turrets could have been removed in the field and that crew could have been left behind. However, the question I am asking is whether they actually did.
I have seen hundreds of pictures of B-29s and, other than the silverplate versions mentioned in the Answer, I have never seen a picture of a regular B-29 (or B-29-A) with guns or turrets removed.
To the contrary, the quote from LeMay in the Answer: (1) indicates that the lower guns were not permanently removed since ammo was provided for them; (2) says that ammo was not provided for the upper guns, which indicates that they were also not permanently removed; and (3) says that the gunners were not removed from the crews.
Thus, the quote from LeMay indicates that, contrary to common belief, the guns were not removed on a permanent basis, if at all. In fact, it may have been nothing more than a proposal that was dropped following strong pushback from the aircrews. Sometimes proposals, if they are controversial enough, get mistaken for facts.

I have no idea, but if you don't get a good answer here you might also try asking on history.SE.

Note that the guns on the B-29 were remotely operated, with the gunners sitting inside the pressure vessel, so you could remove the guns without effecting pressurization.

The older ground crew probably had practice patching holes in fuselages, between flights, so that aspect doesn't pose much problem.

Removing an entire turret, making an aluminum patch to cover the hole, then riveting it in place would not actually be a lot of work for maintenance crews in those days.  They frequently had to patch damage.  They also frequently made changes as operational practice changed (think of the forward firing guns added to 8th army bombers in Europe at the squadron before the factory models had them. )  Removing a turret would reduce the weight.  Also you could leave the gunner behind.  More weight and fewer lives lost if the aircraft is lost.  Late in war there was little resistance.

Yes, in March 1945 Curtis LeMay did order guns and ammo removed, but then shortly after rescinded the order.
Here is a quote from his book titled “Superfortress”:
“At first, we had decided to take out the guns and gunners to make up for the shortage of airplanes, because it would be another factor in helping increase the bombload. However, if the change to low altitude had been quite a step to take, then going in at low altitude with half a crew was something else! I find it impossible to describe the relationship that developed between crew members going into combat together. They were more than a team, they were like a family. If a crew member was sick, or for some reason could not go on a mission, the rest of the crew felt as though part of them was missing, and they would not do as well without him. So flying these missions with half a crew was quite an emotional experience for those who stayed as well as for those who went. Tampering with crew morale was something I did not want to do, and so the gunners went back in, but only with ammunition for the bottom turrets, and with searchlights as the only targets for them to shoot at. I was much more worried about B-29s shooting at each other than I was about any fire we might receive from the ill-equipped and inexperienced Japanese night fighters.“
My guess is they temporarily removed just the guns but left the turrets in place with holes remaining. The holes would not be a problem in the unpressurized turrets.
I think the common misconception/exaggeration you noticed is also attributable to the fact that the “Silverplate” Atomic Bomb B-29s and the later B-29B model aircraft were used in service with defensive guns removed.
I believe those aircraft had the entire turrets removed at the factory, and had flush panels to cover the holes.
Here are some photos of B-29s without the turrets.

Thanks. I had overlooked (or forgotten) that excerpt. Interstingly, like many statements, it seems to incorrectly imply that the B-29s only engaged in low level incendiary raids after Tokyo. Perhaps, especially after the war, LeMay had a vested interest in perpetuating that storyline because he played such a key role in the risky, but successful, decision to "go low".

They might have removed the guns on some Guam planes at least for a little while - since they were under LeMay's watchful eye. But I have seen no evidence that they ever did so on Tinian. As far as I know, any B models that arrived were all sent to Guam. They had further to fly to Japan, so weight would have been more critical to them.


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Did the jet stream really stop the B-29, or was it just bad luck?
The B-29 bomber, used in WWII, was designed as a high altitude bomber, capable of operating from altitudes in excess of 30,000 feet. However, starting with their first missions to Tokyo in Nov 1944, the B-29s encountered a 200+ mph jet stream which made accurate bombing impossible.
After General LeMay arrived in Jan 1945, he reduced their altitude to around 20,000 feet - sometimes as low as 8,000 feet. Because of improved bombing accuracy, higher bomb loads and acceptable losses, the B-29s never returned to 30,000 feet (except for recon and atom bomb missions).
The public perception appears to be that the B-29 could not be used as a high altitude bomber because of the jet stream.  (There were other problems, of course, but the jet stream always seems to be singled out as the reason.)

The picture above has been described as an illustration of the effect of the jet stream on bombing accuracy. See XXI Bomber Command
My understanding of the jet stream (including the jet stream over the US) is that it is stronger in the winter, fairly localized and often changes speed, altitude and location.
Was it just bad luck that the B-29s encountered the jet stream over Tokyo in the winter of 1944? Could the jet stream have moved or weakened by Mar 1945 - when the low level raids were first conducted? Or is there a persistent high speed jet stream over Tokyo or mainland Japan every winter?
ADDENDUM
Vern Haugland of the AP wrote a lengthy article "World's Most Violent Winds Are Problem in Bombing Tokyo" [Evening Star, Washington, DC (Jan 8, 1945), Page A-2] which described, in great detail, problems caused by the winds encountered over Tokyo and the extreme cold at 30,000 feet.
The article did not use the term "jet stream". That term did not appear until 1949 when scientists published studies describing the discovery of "the jet streams of wind blowing from east to west around the world at different latitudes". [Evening Star, Washington, DC (Apr 20, 1949), p. A-6]
Thus, it is possible that, given their limited knowledge of what was happening, the military did not realize that what they were dealing with could have been a temporary phenomenon.
p.s. The picture above is probably not an illustration of problems caused by the jet stream because both the aircraft and the bombs should be traveling at the same speed within the jet stream. Instead, it could be illustrating the effects of wind shear, where winds at different altitudes are moving at different speeds.

The most amazing thing about the B-29 was its defensive fire control. The GE system wasn't just guns pointed by remote control. The gunner aimed at the target with a sighting head, but he was just sending voltages via Selsyn synchros to a central electronic analog computer and the computer did all the actual pointing the guns after correcting for all the variables, like parallax error, lead, ballistics etc. So the sight head's alignment and the gun alignment was never the same. 15 years before the concept started to be applied to flight controls, it was a true "fly-by-wire" control system.

Jets form along warm/cold airmass boundaries, such as the mid latitude Polar and Tropical air mass borders, where the warm side tapers off into a wedge shape at the top of the airmass at the tropopause.  So jets form randomly at different locations along the writhing "belt" of Polar/Tropical frontal boundary that is always present along the mid latitudes around the earth.
It was a bit of bad luck insofar as Japan is at a latitude that places it frequently on the Polar/Tropical airmass boundary if the frontal conditions were right, so jets could frequently form at places that happen to be over a particular target.  Jets were only dimly understood and were likely very difficult to predict with the weather technology of 1945. So the bombardier only knew he was in a jet when he was in it.
Surprisingly, jet stream problems notwithstanding, according to this article, bombing accuracy over Asia was roughly the same as Europe, with about a third of bombs hitting within 1000 ft of the aim point on precision raids (but the Norden was supposed to do better than that).
So I suspect that jet streams were an occasional problem (because they come and go), but the larger one was simply the vertical distance that magnified the Norden's error band, which was a problem in Europe and Asia.  Also, even without being in a jet, if you imagine the variation of winds through 4 or 5 miles of air, and the fact that the Norden made its ballistic wind drift compensation based on the airplane's track/heading offset at its bombing altitude, it's not hard to see the overall problem, jet or no jet.
In Europe, precision was vastly increased when bombing altitudes were lower to the mid teens in the last 6 months, where the Norden really could place bombs with great accuracy (60% hitting within 1000 feet).

That's a great article. Only 4 days of visual bombing per month? The B-29 was equipped with radar which they could use to perform "offset" bombing through the clouds - using the coastlines as reference. But the radar operators with the training to do so would not arrive for several months.

And useless for going after specific targets like manufacturing plants.  It was only good for area bombing.  The WWII equivalent to the precision strike was the B-26 or B-25 bombing from 12-14000 ft where the Norden really excelled.  They are the ones to took out the very small precision targets like small to medium size marshalling yards, bridges etc.

Based on the information you provided, it appears that the jet stream may not have been the primary reason they decided to abandon high altitude bombing - but it provided a convenient excuse. Given that the B-29 and Norden bombsight were the 2nd and 3rd most expensive projects in the War, no one would have wanted to admit that they were not performing as planned. Contemporary newspaper accounts describe the high altitude winds, but not the other issues.

I think the 4 day per month thing might refer to bombable days of specific targets, not Japan overall. There is also some popular misinformation about the Norden.  Its general workings were well known by the Germans during WWII because it was developed in the 30s before the heavy secrecy was imposed. The gap between expectations and results was because it was demonstrated to the Army from only10000 ft.

@JohnK: Why was radar-directed bombing useless for pinpoint targets?  Once you know your location relative to the coastline, it seems that it should be easy (at least in theory) to use that plus the known location of the target relative to the coastline to drop bombs on the target from above the clouds.

Image from ground mapping radar  was too crude for the level of precision required, plus the radar system wasn't integrated into the sight.  You could guesstimate the location of a factory on a radar screen based on shoreline contours, but to pinpoint something a cpl hundred yards across on a little 12 inch (or whatever) screen by eyeball, and to come up with the correct ballistic solution for the release point, forget about it. Only good for area bombing targets. The Norden had a hard enough time with its gryostabilized precision telescope and complex analog computer and autopilot interface.

The B-29 could carry high explosive or incendiary bombs, or a mixture of both.  High explosive bombs have a much higher terminal velocity, there for in the presence of appreciable cross winds, high altitude precision bombing with mixed loads would be nearly impossible.
In a 50 mph cross wind:
HE bomb terminal velocity of 600 mph reaches the ground from 6 miles up in around 40 seconds.  The crosswind would move it around 2/3 of a mile.
Incendiary bombs terminal velocity of 300 mph would take 1 minute 20 seconds with a drift of over a mile.
They weren't going to hit anything and General Le May knew it.  The decision to go low and carpet bomb with incendiary bombs effectively destroyed most of the major cities, but inflicted massive civilian casualties from the firestorms created.

Yes, the incendiary bombs were not streamlined for high altitude. The plan was apparently to drop them at night when it would be safer to bomb from lower altitudes (as the British had done in Europe). The firestorms which occurred at Tokyo, Hamburg and Dresden were not part of the plan. Instead, once again, the weather played the key role in creating them.

Sadly, such a beautiful aircraft tasked with such an ignominious role.

By that time in the war, area bombing using incendiaries was considered a necessary part of the second objective of the bombing strategy - to demoralize the public. [See "The Effects of Strategic Bombing on Japanese Morale" (June 1947), available online] The B-29 was used in other roles, such as strategic bombing of airfields during the Battle of Okinawa to thwart the Kamikaze threat. One of their most successful roles was with the low-level aerial mining missions.

Well, on to better times.  The B-50 did offer some excellent improvements, exceeding 400 mph with 2 added jets.

@RobertDiGiovanni: The jets were only used on the KB-50 tanker version, and only because it couldn't fly fast enough without them to refuel high-speed jet fighters and bombers.  The bomber version of the KB-50 never had them.

The above picture has been described as an illustration of the effect of the jet stream on bombing accuracy.

If the suggestion is that the physical effect of the wind shear around the jet stream upset the flight path of the bombs enough to be visible in a photo taken at close range-- that is highly unlikely and casts doubt on the credibility of any source making that argument.  The jet stream was not the primary obstacle to accurate bombing from high altitudes with the B-29 over Japan.

I've added the link to the picture. I would not normally cite a Wikipedia link as an authoritative source. Instead, I would cite their sources. However, this appears to be a case where the author of the Wikipedia article either engaged in a little editorial license or found the picture on a site which had that caption. In either case, the caption is wrong.


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Matching the Drag Equation to Real Airplane Data
I am trying to compute airplane performance for a piston-driven airplane with constant speed prop using the standard equations for Lift, Thrust and Drag. I have found that there are a couple of useful variables that I can change to make the results more closely match with published data. These are Cd0 (the coefficient of Parasitic Drag) and K, a multiplier used to compute Cdi (the coefficient of Induced Drag).  Cd0 and Cdi are added to together to compute Cd (the total coefficient of Total Drag), which is used to compute the Total Drag Force.
Most of the value of K is given by the formula: K = 1/(Aspect Ratio (AR) * Wing Efficiency (e) * PI). But you can further adjust K by multiplying it by a another constant, which I shall call K0.
You can enter estimated values for Cd0 and K0 and see if the results agree with the published performance data for the aircraft. I generally try to use Sea Level data since that eliminates the variances that arise at different altitudes. I also need to take care to determine the aircraft Weight or Mass that is used in computing the public data.
DATA POINTS
At Maximum Speed (Vmax), Drag should equal Maximum Thrust (Tmax). On a piston or propeller aircraft Tmax decreases with Speed, so you have to use the trial and error method to compute Vmax.
At Maximum Lift/Drag Speed (L/D max), Cdi = Cd0. This is the low point for Cd and Total Drag. There are equations for computing the Speed at L/D max.
At the Best Rate of Climb Speed (Vy) for a piston-driven airplane with a constant speed prop, Cdi = 3 * Cd0. There are equations for computing the Speed at Vy speed.
My understanding is that you must use the trial and error method to determine the Best Angle of Climb Speed (Vx) for a piston-driven airplane with a constant speed prop and I am not aware of any necessary relationship between Cdi and Cd0.
At Takeoff Speed, the coefficient of Lift (Cl) = Clmax. I am not aware of any necessary relationship between Cdi and Cd0.
METHODOLOGY
I create an Excel spreadsheet which has a graph like this:

I initially leave K0 at 1 and insert different values for Cd0 until I find one where the lines for Tmax and Total Drag cross at the published Vmax.
I determine whether the lines for Parasitic Drag and Induced Drag cross at the published speed for C/L max. If not, I adjust Cd0 and K0 until they do, while not changing the computed Maximum Speed.
I determine whether, at published Vy, Cdi = 3 * Cd0. If not, I adjust Cd0 and K0 to try to create the best fit.
I can change the Thrust and Drag values a bit by changing Prop Efficiency and Wing Efficiency, respectively.
Given that the Drag Equation follows a fairly fixed plot, there may be situations where I get two of the data points to fit perfectly, but a third data point does not - which means that I must use values which generate the "best fit".
The computations in the chart are for an FM-2 Wildcat, which has a top speed of 300 mph and a Vy of 144 mph. I used value of .01825 for Cd0 and 2.74 for K0. The results match the published data almost perfectly (but I only found 2 data points so that is inevitable).
I am not taking including propeller efficiency (Np) into account. I expect that the propeller efficiency should reach the max value before takeoff speed and will remain at that max value until just before Vmax. So, it might reduce the actual Vmax from what I have computed.
ADDENDUM
From another I learned that "The ratio between max range and max endurance speeds is always 3^0.25." Since the Max Range Speed equals the L/D max Speed, you compute Max Endurance Speed by dividing L/D max Speed by 3^.25. It turns out that Max Endurance Speed equals the Vy Speed. So this provides a mathematical connection between the two.
Also, reviewing the effect of changes in altitude, it appears that for both Vy and L/D max, the Cl and the relationship between Cdi and Cd0 remain unchanged as altitude increases. And Vy Speed remains equal to L/D max Speed divided by 3^.25.
QUESTIONS

While not perfect, does this methodology make sense?

Are there other data points I can use?

Are there any other speeds, besides L/D max and Vy where there is a fixed relationship between Cdi and Cd0?

What you describe is commonly called the quadratic drag equation.

While not perfect, does this methodology make sense?

Yes, see answers here, here and here, for example. But you can easily refine the results. It helps to adjust zero-lift drag with speed, for example, as done in this answer. This should give better results than your approach with K0. Your zero-lift drag coefficient is almost certainly too low.

Are there other data points I can use?

Yes, anywhere you can determine drag precisely. This is harder than it sounds, however. Normally you will only get a combination of drag and thrust. Once thrust is known exactly, the drag coefficient can be calculated if you know the airplane weight, load factor, flight speed, rate of acceleration, climb speed and the atmospheric conditions for this data point.

Are there any other speeds, besides L/D max and Vy where there is a fixed relationship between Cdi and Cd0?

No, not the way you desire.
I do not know how you calculate thrust, but I am sure you underestimate the drag increase due to prop wash. Here is an answer with a link to a very interesting article on the subject. Also I wonder whether you included the effect of constant CAS on climb speed. Close to stall speed the quadratic drag equation loses some of its shine due to separated flow beginning to nonlinearly increase drag. Here it is best to assume a linear lift curve slope when calculating the induced drag coefficient but to gradually reduce the lift coefficient you use for speed calculations as stall speed is approached.


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Computing an Energy-Maneuverability (E-M) Diagram for an Aircraft
Computing an Energy-Maneuverability (E-M) Diagram for an Aircraft
The Energy-Maneuverability (E-M) Diagram is a helpful way to show the horizontal turning performance of an aircraft at various speeds.
While I have found many discussions of these diagrams online, I have not found anything explaining how to compute a theoretical diagram for a specific aircraft.  Here is my attempt to create one for a WWII fighter aircraft - the FM2 Wildcat:

QUESTION
Am doing this right?
To simplify discussion, I have listed the Variables in Appendix 1 and the Equations in Appendix 2.
MAXIMUM ANGLE OF ATTACK (AOA) LINE
The left blue line (currently labelled AoAmax and partly overwritten by the green line) shows the turn rates at maximum Angle of Attack (AOA) at various speeds.
Simply put, the Angle of Attack (AOA) produces Lift. The equation for Lift max is:
  Lift = cL*Q*S [eq. 2.0]

where cL is the coefficient of Lift, Q is the Dynamic Pressure and S is the Wing Area. For a given speed, the maximum cL produces the maximum Lift. The cL is a function of the Angle of Attack. The AOA is "the angle at which the chord of an aircraft's wing meets the relative wind. The chord is a straight line from the leading edge to the trailing edge". The maximum AOA is generally around 16 degrees. The cL is generally around 10% of the AOA.
At take-off speed, Lift is just greater than Weight, so the aircraft can fly. But the aircraft is unable to turn because the aircraft needs all of the lift to fly.  As aircraft speed increases, lift at maximum AOA increases, which means that there is more lift available for turning.
To create this line, I started by computing the take-off speed (Vto). At this speed, Lift max = W (Weight) and the turn rate is zero. The equation for Lift max is:
  Lift max = cLmax*Q*S [eq. 2.1]

The equation for Q (Dynamic Pressure) is:
  Q = 1/2*ρ*V^2 [eq. 4.0]

We can combine the above into
  W = Lift max = (1/2*ρ*V^2)*(cLmax*S)

Solving for V gives:
  Vto = SQRT(W/(1/2*ρ*cLmax*S)) [5.1]
  Vto = SQRT(7048/(1/2*0.00237*1.6*260)) = 119.57 ft/sec = 81.53 mph

You can then compute the AOA max speed for each different G-Force (GF). At each different GF:
  Lift = GF*W

Since everything else remains constant, you can modify the Vto equation to:
  Vgf = SQRT((GF*W)/(.5*ρ*cLmax*S))

Since all of the variables remain unchanged, except for GF, you can simplify this to:
  Vgf = Vto*SQRT(GF)

Thus, where GFmax = 7:
  Vgf = 119.57*SQRT(7) = 316.36 ft/sec = 215.69 mph.

You might notice that this is the same equation used to compute speed for accelerated stalls. Thus, for each GF, the Vmin value on the E-M Diagram equals the minimum speed for an accelerated stall.
Since the E-M diagram computes horizontal turn rate and since the maximum horizontal turn is in level flight, you can use the standard equations to compute the turn rate at each Vgf:
  BankAngle = DEGREES(ACOS(1/GF)) [eq. 6.0]
  TurnRate = DEGREES(TAN(RADIANS(BankAngle)))*g/Vgf [eq. 7.0]

Thus where GF = 7:
  BankAngle = DEGREES(ACOS(1/7)) = 81.79 deg
  TurnRate = DEGREES(TAN(RADIANS(81.79)))*32.174/316.36 = 40.39 deg/sec

This means that the shape of the Vmin curve will be about the same for all aircraft.  For a given aircraft, the size of the curve will vary depending on S (wing area) and standard W (weight).  For a given flight, the size will vary depending on actual W and p (air density).
The curve will extend from Vto (where TurnRate is zero) to the speed and turn rate at maximum GF.
G-FORCES (GF) LINES
The red lines show the G-Forces at various speeds.  You can use the equations above to plot these lines.
In level flight, each G-Force (GF) has a corresponding BankAngle, which remains constant for the entire speed range. You can compute this BankAngle using the BankAngle equation (above):
  BankAngle = DEGREES(ACOS(1/GF)) [eq. 6.1]

You can then use TurnRate equation (above) to compute the TurnRate at various speeds:
  TurnRate = DEGREES(TAN(RADIANS(BankAngle)))*g/V [eq. 7.0]

Ideally, the top GF Line will show the upper boundaries of what the aircraft and pilot can tolerate.  For modern aircraft, this line is generally at 9 g's, which is what the pilot can tolerate on a sustained basis.  Those aircraft can generally operate near this upper limit in level flight.
I am using a lower GF Line for the FM2 because the aircraft simply does not have the power to reach this limit in a sustained turn and rarely (if ever) in an instantaneous turn.  (Anecdotally, a test pilot who dove the plane straight down was never able to pull more than 7.2 g's.)
VNE LINE
Normally, as noted below, there should a vertical line at Vne (the Never-Exceed Speed). However, the Wildcat was one of the extremely rare aircraft for which there was no real Vne. You could apparently dive the aircraft all the way to terminal speed (around 550 mph).
SUSTAINED TURN (PS) CURVES
The green line (labeled PS = 0) shows the maximum possible sustained TurnRates at various speeds in level flight.
The first three lines: the Max AOA Line, the Max GF Line and the VNE Line should set the boundaries within which the aircraft can operate.  The Sustained Turn Lines show the TurnRates available on a sustained basis within those boundaries.
The maximum TurnRate is determined by a combination of Excess AOA and Excess Net Thrust above what is required for straight and level flight.  Where the Line matches the Max AOA Line, the TurnRate is limited by the AOA Available. There is still Excess Net Thrust Available, which can be used to accelerate.  At the peak of the Line, where the TurnRate is greatest, the aircraft is using all of the available AOA and Net Thrust. At higher speeds, the TurnRate is limited by Net Thrust Available. The reason is that, as speed increases, Thrust decreases (on a prop aircraft) and Dp increases to the point that the excess of Thrust over Dp is not enough to offset the Di at the maximum BankAngle for that speed. This means that BankAngle and TurnRate must decrease so that Thrust = Total Drag (Dt). At the top speed all available Thrust is required to offset Dt in level flight. There is no excess Thrust to handle the increase in Di associated with a turn.
Since the max TurnRate is computed assuming we are using all excess Thrust to increase TurnRate, there is no excess Thrust remaining to increase speed. Thus, to increase speed, you must decrease TurnRate.
The computation of the maximum TurnRate for different speeds involves several equations and takes several steps (or, as below, you could combine them all into a single equation).
(1) Compute the maximum Di available using:
  Di= Thrust max - Dp
  where:
    Thrust max = 550*BHPmax*η/V [eq. 1.0]
    Dp = cD0*Q*S [eq. 3.1]

(2) Compute the maximum cDi using:
  cDi = Di/(Q*S) [eq. 3.2.1]

(3) Compute the related cL using:
  cL = SQRT(cDi*PI()*e*AR) [eq. 2.1]

If the cL is greater than cLmax, then cL = cL max and there is no reduction in max TurnRate.
(4) Compute the maximum BankAngle at that speed using:
  BankAngle=DEGREES(ACOS(W/Lift)) [eq. 6.1]
  where:
    Lift = cL*Q*S [eq. 2.0]

(5) Compute the TurnRate for that speed using:
  TurnRate = DEGREES(TAN(RADIANS(BankAngle)))*g/V [7.0]

You can also compute the PS curves for situations where you climb or descend. You can descend to use gravity to supplement Thrust and increase TurnRate. The resulting curve line will be higher than the PS0 curve. Conversely, climbing will decrease TurnRate.  The resulting curve line will be lower than the PS0 curve. However, to keep things simple, I am just computing the PS0 curve.
Several posts below discuss how to compute the PS curve in greater detail. As those discussions indicate, those computations are made using various assumptions, some of which are necessary to simplify the computation. This is an area where one would expect an E-M diagram based on computations to vary from an E-M diagram based on actual flight data.
CONTOUR MAPPING (added Jan 17, 2023)
Several of the responses below refer to contour graphing, which is a new concept to me. Although, I am familiar with contour maps, but never considered how those contour lines were created. For others who are in the same boat, here is a simple explanation of what that involves.
In the question above, I am plotting lines created using specific formulae. In contrast, contour mapping involves creating arrays of data and then calling a contour-plotting formula which will parse the data, perform some interpolations and plot the data.
For example, the MatLab program creates several different "arrays" of data indexed by reference to increments of Speed and GLoad. For example, the PS value for Speed=300fps and GLoad=5 is stored in a PS array at location PS[50Speed_Increment,5GLoad_Increment]. The program then calls a built-in contour graphing function in which you specify the values that you want plotted (e.g. PS=0). The program will then parse the data to estimate where PS would equal 0. It correlates GLoad to TurnRate and plots the Speed and TurnRate values on the chart.
My research indicates that Excel contour charts are unable to draw the kind of contour charts shown below. One major limitation is that Excel apparently cannot create a single chart that uses more than one data source, e.g. GForce, cLmax and PS.
APPENDIX 1. VARIABLES
AR = Aspect Ratio of Wing = 5.55
cL = coefficient of Lift; cLmax = 1.6
cD = coefficient of D = sum of cD0 and cDi
cD0 = coefficient of Dp = 0.021
cDi = coefficient of Di (equations below)
D = Drag
Dp = Parasitic Drag (equation below)
Di = Induced Drag (equation below)
e = Wing Efficiency = 0.75
BHP = Brake Horsepower; BHPmax = 1300
g = Gravity = 32.174 ft/sec^2
GF = G-Force
η = Propeller Efficiency = 0.75
ρ = Air Density at Sea Level = 0.002377 slug/ft^3
Q = Dynamic Pressure (equation below)
S = Wing Area = 260 ft^2
V = Velocity or Speed (ft/sec)
Vgf = Speed for Given G-Force
Vmin = Minimum Speed for Given Turn Rate
Vto = Takeoff Speed (equation below)
W = Weight = 7048 lbs

APPENDIX 2. EQUATIONS
[1.0] Thrust = 550*BHP*η/V
      Thrust max = 550*BHPmax*η/V
[2.0] Lift = cL*Q*S = (1/2*ρ*V^2)*(cL*S)
      Lift max = cLmax*Q*S = (1/2*ρ*V^2)*(cL*S)
[2.1] cL = SQRT(cDi*PI()*e*AR) (from equation 3.3.1)
[3.0] Drag (D) = Dp+Di = cDt*Q*S
[3.1] Parasitic Drag (Dp) = cD0*Q*S
[3.2] Induced Drag (Di) = cDi*Q*S
[3.2.1] cDi = Di/(Q*S)
[3.3.1] cDi = cL^2/(PI()*e*AR)
[4.0] Dynamic Pressure (Q) = 1/2*ρ*V^2
[5.0] V = SQRT(W/(.5*ρ*cL*S))
      Vto = SQRT(W/(.5*ρ*cLmax*S))
[5.2] Vgf = SQRT((GF*W)/(.5*ρ*cLmax*S))
[6.0] BankAngle = DEGREES(ACOS(1/GF))
[6.1] BankAngle = DEGREES(ACOS(W/Lift))
[7.0] TurnRate = DEGREES(TAN(RADIANS((BankAngle)))*g/V

One aspect of an EM chart is that you don't just want to plot Ps=0, but also other contours of Ps.  This can be done algebraically (like you have done) for simple drag polars and engine models, but it is a real pain.  Instead, it is normal to use a contour plot to establish these lines implicitly.  I.e. calculate Ps for every point on a grid of turn rate and velocity, then contour that data.  Excel's contour plots are terrible, you'll want to use Matlab or Python.

The interesting aspect here is you get an airframe EM curve readily, but it is power (thrust) constrained.  The EM diagrams have lines where they say "you can do this but you are gaining or losing energy (speed or altitude)" based on available thrust and drag.  (Diving adds thrust).  Sustained turning would be no  net energy loss in the manuver.  Naturally, more drag from turning requires more thrust.  No wonder best rate of turn is around Vy.

@Rob, Yes I can add some of those lines. I was just concerned that the diagram would get a bit messy, especially since the results seemed to simply parallel the Ps = 0 line. To compute those numbers for a particular vertical speed (VrtSpd), I determined the angle required to generate the required climb or descent (Angle=DEGREES(TAN(VrtSpd/Spd)) and then determined the portion of the Gravity force to add or subtract from Di (Force=W*SIN(RADIANS(Angle)).

@PhilCrowther well the key would be to have the lines for both planes.  I recently saw an EM for a Sabre and Mig 15.  The Sabre had automatic slats (like Me109), which helped its low speed ROT.  The Mig climbed very well.  Power/Weight will factor in.

@PhilCrowther I would argue that without different lines of Ps, it isn't really an EM chart.  The lines of Ps are what let you compare two aircraft at a particular point and determine who has the advantage.  The greater point is the power of using contour plots instead of algebraically solving for the curves.  This is a must-do when the drag polar or engine model are more complex.  It is also a lot easier.

@Rob I have replaced the diagram with one that has different Ps lines. To me, they don't seem that helpful because they are completely parallel to the Ps0 line. Thus, in comparing E-M diagrams between airplanes, I would think that you would get a decent idea of relative performance by simply comparing the max values on the Ps0 line. However, this may be due to the fact that I am using a very simple flight model and not using things like drag polars and contour plots. Do you have any good cites to discussions of those topics?

@Rob, I have replaced the previous diagram with one that shows the Ps lines using feet per second rather than feet per minute and have switched to a bigger chart that has room to plot things like turn radius.  Let me know if you think the print is too small.

@Robert The MiG was faster and performed better at higher altitudes than the Sabre. Presumably that would show up in side-by-side E-M Diagrams for the planes at different altitudes. So the goal of Sabre pilots was to get the MiGs to come down and play - like the Zero pilots tried to do with Allied pilots. The MiG was an unpleasant surprise, especially for B-29 crews.

I call those diagrams turn rate diagrams, because that is plotted over airspeed. Never mind.

Am doing this right?

I think so, your result looks fine. Now let's start with the explanation  how to compute a theoretical diagram for a specific aircraft which you are missing on the interwebs. Please allow me to use the standard symbols in my equations which differ from what you used.
First you draw lines of equal turn radii which radiate away from the origin. Next, you draw lines of equal load factor which are hyperbolic. This is what you get:

The equations for those lines are $$\text{turn rate}=\frac{\text{speed}}{\text{radius}}$$ and $$\text{turn rate}=\frac{g\cdot\sqrt{n_z^2-cos\gamma}}{\text{radius}}$$
$n_z$ is the load factor and $\gamma$ is the flight path angle, positive values of which denote a dive. The plot above is for horizontal flight, so $cos\gamma$ = 1. This is to account for the vertical component of lift.
You will notice that I used odd load factors below $n_z$ = 2. They are chosen for specific bank angles, namely $\Phi$ = 30° and 45°.
There are two ways to continue, one for instationary turns and one for stationary turns. Stationary means you can fly this without losing altitude, whereas instationary turns trade height for additional energy to maximize the turn rate. Of course, you could as well calculate a climbing turn where less energy is available for turning.
Instationary turns are easy: You plot two segments, one for minimum speed and one for maximum load factor. The turn rate at minimum speed is determined by the wing loading and the maximum lift coefficient $c_{L_{max}}$ and is $$\text{turn rate}=\sqrt{\frac{(n_z^2-cos\gamma)\cdot g\cdot c_{L_{max}}\cdot S\cdot\rho}{2\cdot m\cdot n_z}}$$
Expressed over speed, this becomes $$\text{turn rate}=\frac{\text{speed}}{R_{min}}=\frac{g\cdot\sqrt{\left(\frac{\rho\cdot\text{speed}^2\cdot S\cdot c_{L_{max}}}{2\cdot m\cdot g}\right)^2-1}}{\text{speed}}$$
with $m$ the airplane mass, $S$ the wing area and $\rho$ air density. When you vary $n_z$ (upper equation) or speed (lower equation), you get the line for maximum lift. If the term in the square root is negative, speed is lower than minimum speed at 1 g and no solution is possible.
The next limit is the maximum load factor the structure can withstand and applies above the minimum speed. The result should look like this:

where I added coloured lines for different engines with different thrust over speed behavior. A piston engine is represented by the blue line. Above the coloured lines the airplane cannot sustain the turn rate with engine power alone while below it the airplane needs to throttle the engine in order to not climb away. In other words: The coloured lines are the limits for stationary turns.
You will note that this is not for the Wildcat but some generic plane. Also, $c_{L_{max}}$ is usually not constant but goes up with load factor because of the increasing Reynolds number. To do this correctly for a specific airplane will need to include all speed-dependent adjustments.
If your airplane is limited in angle of attack (like the Phantom II was), replace $c_{L_{max}}$ by $c_{L\alpha}\cdot\alpha_{max}$.

This is an excellent explanation and expansion of my question. To assist readers, I have tried to change my variables to match yours.  I did not change GF to n because I thought there might be some confusion with the Greek "n" used to represent prop efficiency.

@PhilCrowther The Greek $\eta$ looks distinct from an n, having that longer right leg.

Those are called EM diagrams, not just turn rate chart because of the inclusion of Specific Excess Power (SEP/Ps) curves, which can be calculated by (Thrust - Drag) * velocity / Weight, where Thrust should be the thrust component in the stability axes.

Generally, these diagrams are produced from extensive flight test and data collection. You cannot get sufficient accuracy from the simplified equations you are using. Computational Fluid Dynamics applications software use linear approximations of Navier Stokes equations and mathematical models of the complete airframe and wing to perform an iterative (millisecond by millesecond) computation, which still needs to be validated by flight test data.
That is not to say that the approach you are using cannot generate a qualitatively accurate picture. I think it does. I would just be hesitant to put any actual values on the graph that imply accuracy that it does not have.
But just looking at your diagram, here are a few comments.

There should be a vertical line at Vne.
The left boundary is not a minimum speed line. It is a maximum Angle of Attack (AOA) line. A 1G line would reflect level flight. The 1 G line, where it intersects with the far left (Max AOA) line would designate 1 G stall speed, or minimum Takeoff speed. The Max AOA line would continue down and left of that point to the zero G line.
It is important to understand that Specific power is about sustainability, not instantaneously capability. The zero Ps line should not be aligned at the left (AOA) boundary at very high Gs. Can FM-2 Wildcat really sustain (continuous) 3 Gs at Maximum AOA (Clmax)? The zero Ps line should intersect the far left edge of the envelope at the point you would end up at if you select full power and pull on the pole to just below stall AOA.
In your text, you say
`

The green line shows the Specific Power (PS) curve for full power in
level flight (PS0). This shows the maximum possible TurnRate at
various speeds

This is not correct. First, it shows the zero Specific excess power line at each airspeed, specifying the maximum sustainable G-load at that airspeed. It does not show level flight. That is simply the 1G line. Secondly, it is not the maximum possible turn rate at that airspeed. It is the maximum sustainable turn rate at that airspeed. I emphasize this because this is the essence of the entire concept of Energy maneuverability. It is about sustainability, not about instantaneous capability.

"The left boundary is not a minimum speed line. It is a maximum Angle of Attack (AOA) line" only when your airplane is AoA limited, like the Phantom II. Normally it is indeed a minimum speed line at the respective load factor.

I agree with all of your comments and will revise my explanations. Re #1, I have read that the design of the Wildcat made it impossible to exceed Vne, even when pointed straight down. Re#2, yes, that line represents a max AoA line. I will relabel. Re#3 and #3, yes, that line is supposed to show the max sustainable turn in level (vertical) flight

@Peter,  All aircraft are AOA limited. It's called Cl max or stall Angle of Attack.  Minimum speed at at some specific load factor is just another way of looking at it, except that using that perspective does not take into consideration gross weight. (The less you weigh, the higher the allowable load factor for the same stress on the airframe structure).  But Max AOA is the same, regardless of gross weight, bank angle, or airspeed.

@Phil, I found your statement about Vne surprising.  Do you know what it's terminal velocity was when pointed straight down with full throttle? Do you know- was this design constraint you mention accidental or deliberate?

@Charles, According to the Pilot Handbooks, you could point the aircraft straight down and (apparently) the airframe and propeller would create enough drag to prevent airframe damage. The only concern was to make sure that the engine RPM did not exceed 3050 [F4F, p. 61] or 3100 [FM2, p. 36]. The terminal velocity was estimated at 475 kts [F4F] or 370 kts at 15k feet [FM2].

"But Max AOA is the same, regardless of gross weight, bank angle, or airspeed." Right only when you are AoA limited, but wrong for most airplanes. In all airplanes the maximum lift coefficient and maximum AoA go up with speed due to Reynolds number effects. This becomes even more pronounced for high pitch rates.

@Peter, Thanks for that additional bit of information. To improve my computation, I should also use the NACA wing data for the FM2. At the speeds I am using, would the Reynolds number have much effect? In any case, this sounds like something I should add to the computation for all those people who want to make the computations for faster aircraft.

@Robert, you noted that be level sustained turn rate seems at Vy. I hadn't noticed that, but you are correct. In a separate exercise, I had calculated that Vy for the FM2 is 138 mph. It could be that I got to the same results using the same formula or a variation of the same formulae, but it is something I had not noticed.

@PhilCrowther This answer shows a large collection of windtunnel data for maximum lift coefficient over Re. Above 2 million you can see a fairly steady increase over log(Re) which should also be valid for the Wildcat wing.

@Charles, For the record, I agree with your initial statement that my diagram is not an accurate depiction of aircraft performance - it is merely an approximation "not to be used for navigation". For example if you look at the many F-16 E-M Diagrams available online (and presumably based on actual flight data), the lines for climbing turns do not start along what appears to be the max AoA line. Also, in my first comment above, there were a couple of typos - #3 and #3 should be #3 and #4 and "(vertical)" should be "(horizontal)".

@Peter, do you know how much (percentages) the Maximum AOA and lift coefficient increase with speed increases, and over what range of airspeeds do these increases occur? I ask because if these are significant, much of what I believe is extremely important about AOA is simply wrong.

@CharlesBretana here you will find c${L{max}} data over Re for many different airfoils and NACA TN 2525 presents data for maximum lift over pitch rate.

@CharlesBretana: Another factor is the local Mach number which can become important at flight Mach numbers as low as 0.3. Not all airplanes have an AoA limiter which restricts AoA to less than 23°.

Okay, in reading this data, it does not explain how much Reynolds number can change within the accesible flight envelope of general aviation aircraft. How significant are the changes in the maximum aeridynamically attainable AOA caused by the factors you are referencing? If the max AOA increases or decreases by only 1% or 2 %, then it is not significant, and any such effects are not germane to a discussion about AOA and its importance. How large are these effects?

Secondly, even if these effects are significant, they only change the maximum attainable AOA, they do not eliminate it. Every aircraft still has a maximum attainable AOA. It might be determined by when the aircraft wing stalls, or by when the Lift Coeeficient starts to decrease (instead of increasing), as AOA increases, or by when directional control is lost, or some other factor, but it always happens.

@CharlesBretana Realistic pitch rates increase c${L{max}}$ by as much as 50%. Is that enough to be significant? Also, given the jump from 1.2 at Re=10$^6$ to 1.5 at Re=10$^7$ for a typical airfoil is also quite significant, but less than what pitch rate is able to do.

With all due respect, "Realistic pitch rates increase cLmax by as much as 50%". I'm not asking about Pitch Rates. I'm questioning your assertion or implication that some aircraft do not have a limit on attainable AOA. What does the quoted  statement mean, anyway? Are you talking about rotational inertia, where you can create a pitch rate that overshoots normal max AOA from sheer inertia? Jeez, what about a spin, the F-4 can spin. Does that mean it's max AOA is way higher, cause it goes up in a spin?

And I don't see how a higher pitch rate can increase the lift coefficient by 50%. Realistic attainable AOAs are already 25-35 degrees. As they get higher the full aerodynamic force vector tilts further and further backwards from the direction normal to the flight path vector. Lift, therefore is decreased by that effect, significantly, as AOA increases. Yet you say it increases by 50%, just because we get there faster (pitch rate)?

I wrote a simple turn code in Matlab.
Here is the EM / Doghouse plot at sea level.

It looks like it agrees with your chart.  The reason we don't see the Ps curves 'close' inside the CLmax line is insufficient thrust.  If you increase the engine power, you should see the center of the 'eye' move to the right.
Moving to a non-constant propeller efficiency model would also be a nice update.
I've calculated Ps to the left of the stall line -- but that should be taken with a heap of salt.  Not only will the wing not provide enough lift to turn at those points, there would be tremendous drag that is not accounted for in the simple parabolic polar.
I think you will see how using a contour algorithm to draw the lines greatly simplifies the code.  All of the equations are in their canonical 'forward' mode.  In addition to eliminating all the algebra required to put things into other forms, this approach also enables more complex models that can not be algebraically re-arranged -- such as a complex drag polar or thrust model.
Yes there are a lot of lines -- the chart can be improved by labeling the contours less frequently and less conspicuously.  In addition, use of different line weights, styles, and shades of grey / colors can help a lot.
In general, I find that complex plots like this are still usable/readable as long as the various lines are crossing one another.  If they run parallel, then the different trends become very confusing to discern.
clear all
format compact

Sref = 260;
W = 7000;

% Limit load factor.
nmax = 7;

% Maximum speed to be considered.
Vmax = 500; % ft/s

etap = 0.75;
Pmax = 1300 * 550; % lbf ft / s

CLmax = 1.6;

% Simple parabolic drag polar.
AR = 5.5;
CD0 = 0.021;
e = 0.75;

% Only perform calculations at sea level.
rho = 0.0023769;

g = 32.1740; % ft/s^2

% Resolution to perform calculations, increase for smoother lines.
nn = 50;
nV = 51;

ns = linspace( 1, nmax, nn );
Vs = linspace( 1, Vmax, nV );

for in = 1:nn
    n = ns( in );

    for iV = 1:nV
        V = Vs( iV );

        q = 0.5 * rho * V^2;

        L = n * W;

        CL = L / ( q * Sref );

        CD = CD0 + ( CL^2 ) / (pi * e * AR );

        D = CD * q * Sref;

        T = Pmax * etap / V;

        Ps = V * ( T - D ) / W;

        omega = g * sqrt( n^2 - 1.0 ) / V;

        nmat( in, iV ) = n;
        Vmat( in, iV ) = V;
        Psmat( in, iV ) = Ps;
        CLmat( in, iV ) = CL;
        omegamat( in, iV ) = omega * 180.0 / pi;
    end
end

figure( 1 )

[c,h] = contour( Vmat, omegamat, nmat, 1:1:nmax, 'k' );

axis( [0 Vmax 0 50] )

clabel( c, h );
hold on

[c,h] = contour( Vmat, omegamat, CLmat, CLmax * [1 1], 'k' );
clabel( c, h );

[c,h] = contour( Vmat, omegamat, Psmat, 0:50:1000, 'k' );
clabel( c, h );
hold off

xlabel( 'V (ft/s)' )
ylabel( '\omega (deg/s)' )

Since contour mapping is new to me, I have added a section to my question where I try to explain what it means. Let me know if my explanation is correct.

@PhilCrowther I think your understanding of contour plotting is pretty much correct.  I've created a simple contour plot in Excel, but it is very awkward and I haven't found much capability to adjust things.  There probably is a way, but I would recommend using a tool with better plotting capability (Matlab, Octave, Python/Matplotlib, Gnuplot, TecPlot, etc.)

Here is a sample EM diagram for an F-16.

Note that we do not expect all of the Ps lines to intersect the CLmax limit.
It looks to me like you may have a problem with your induced drag term not being sensitive to increased CL required for the turns.

That was what I was seeing.  I was finally able to get my computations to create little pops up at a climb rate of 58 and 59 ft/sec. Part of the problem is that the FM2 does not have anywhere near the thrust of the F-16. And it could be that I need to modify my formula for computing cDi. I am using a "standard" formula that uses cL as an input. However, I seem to recall that this formula just yields an approximation of cDi.  I found some data about the NAVA 23015 airfoil online at airfoiltools.com.  I will see how closely my formula approximates the results shown there.

CDi as a function of CL is OK -- but the CL needs to be CL=nW/(qS) taking into account n - the load factor during the turn.

Yes, the PS computation takes the load factor into account when computing allowable BankAngle.  For example, at the speed of 150 mph on the line for PS+50, I computed that you need cL = 0.4711 for level flight and a ClimbAngle of 13.251 deg to climb 50 ft/sec. Using the formulae for Thrust, Dp and W (where allocable W = W*SIN(RADIANS(ClimbAngle)), I computed you could have a max Di of 508 ft_lb, equivalent to a cDi of 0.339 and a cL of 0.6665 (which is about 1.414 of the cL for level flight). I then computed an allowable flight angle of 45.021 (which corresponds to that load of 1.414.).

So the problem must be lack of Thrust. At 150 mph, the thrust for the FM2 is 4,438 ft_lb. Since an F-16 can climb vertically, the F16 thrust must be in excess of the gross weight of 23,768.

Out of curiosity, why does the F-16 chart have a line for a load factor of 1?  Isn't that the load factor for straight and level flight, with no turn?  Perhaps I am missing something there?

In climbing & turning flight, L=nWcos(theta).  You appear to be using a sin instead of cos.
When computing Ps, we usually ignore the cos(theta) in the formula for L (and thereby CL).  With it, the calculation becomes iterative -- the climb angle depends on the drag, drag depends on the lift, lift depends on the climb angle.  While this seems contradictory when you derive Ps from forces, it makes more sense when you work from an energy perspective.  A pilot can choose to put the excess power into climb or acceleration (or some of both).  The angle is not determined.

I agree that the 1g line should correspond to zero turn rate.  Not sure what is going on here.  This figure was taken from a F-16 SAC chart -- prepared by GD and delivered to the USAF.  It is certainly a reputable source.

I checked several other SAC charts -- for the F16 and other aircraft.  This one was the only one with an EM diagram.  The n=1 contour is absolutely wrong.  Good eye.  All I can figure is some idiot manager thought the figure looked 'bad' without a n=1 line and told some poor draftsperson to insert one.  Somehow it made it out the door.

You are correct regarding cos(theta). In fact, when creating the chart, I accidentally omitted cos(theta) from my equation, thinking it was a vestige from the days when I thought you had to increase Lift to climb (when I did use sin(theta)). I had forgotten that, as pitch increases or decreases, the required Lift decreases (and required Thrust increases or decreases).  Both you and @Peter (and maybe everyone else) were pointing this out and I missed it. It does not affect my initial discussion of PS=0, but should affect the plot lines where PS<>0 by slightly increasing the available Turn Rate.

This discussion also highlights another issue that I have wondered about. When computing horizontal turn rate where PS<>0, do people take into account the effect that pitch has on BankAngle relative to the horizon?  For example, if your BankAngle in level flight is 60 degrees and you pitch up (or down) by 30 degrees, your BankAngle relative to the horizon is now 63.43. Is this something that people take into account (along with the fact that the Lift vector, when viewed from above, is no longer perpendicular to the direction of flight)?  Or are these effects too small to worry about?

When generating performance charts (flight envelopes, skymaps, doghouse charts, etc.) (i.e. when computing Ps) we assume theta=0.  Ps is not simply rate of climb.  It is the rate of change in energy height (ze=h+0.5*V^2/g).  You can put Ps into altitude or velocity (or both).  At Ps=0, you can trade altitude for speed.  At Ps<0, you can dive or slow down.

@RobMcDonald This EM chart has a positive turn rate with load factor = 1 is simply because it is using normal load factor Nz, which is in the body Z axis.

I might add that for a propeller aircraft, all of the Ps lines intersects at the CLmax is not necessarily wrong, because the thrust generated by the propeller reduces as speed increases, as opposed to that in an F-16, that the subsonic thrust increases with mach number. This affects how Ps curve is shaped and where Ps is culminated.

Here is a simple contour plot example done in Excel...  It is a form of a 3D surface plot -- where your viewpoint is from directly above.

The red column ends up being the X-data.
The blue row ends up being the Y-data.
These can probably be swapped/transposed somehow.
The green data is a 'matrix'.  Each cell contains a computation corresponding to the X,Y of the matching row,column.
In this case, the equation is:
=(F$1-0.5)^2+($A6-1)^2-(0.5)^2

I.e. f(x,y)=0 will be the equation of a circle centered at 0.5,1 with radius 0.5.
The fact that Excel draws the grid makes it particularly easy to understand how contouring works.  Look at an individual square with a contour line drawn through it like this one...

Each of the corners has a value associated with it.  This is the value from the green data in the matrix.
In this case, we are interested in drawing a contour at zero -- but we could draw any other value.  We start by searching for any edge that has endpoint values that differ in sign -- i.e. one is greater than zero and the other is less than zero.
Whenever we find such an edge, we perform interpolation to create a point along that edge that estimates where the zero value would be.  Then for each square, we connect those interpolated points together.
We keep doing this for the entire matrix and the result is a contour plot.
A Mathematician might call this an implicit function.

I won't suggest using bank angle to calculate turn rate. The EM charts are for all bank angles, and is not limited to horizontal turn situations. However the calculations are usually done with an assumption of bank angle = 0, and then be cross checked with flight test data.
In my program for calculating F-16 EM diagrams shown below, I'm simply using:

Turn Rate = gravity * sqrt((Load factor^2) - 1) / Velocity

Also be careful about which load factor you're calculating, as they're in different axes systems, detailed in another answer at https://aviation.stackexchange.com/a/105029/71947
For an EM diagram, you also need the calculation of Specific Excess Power (SEP/Ps):

Ps = (Thrust*cos(α) - Drag) * velocity / Weight

Thrust is multiplied by cos(α) because Thrust Force is in the body axes, and we must cast it onto the stability axes, as the Drag Force is along the stability X axis.

Now here I took a step further to calculate the F-16 EM chart without an AOA limiter:


text:
What is a shockwave, and how does it form?
so I came across this picture on this website: https://howthingsfly.si.edu/aerodynamics/shock-waves

I’m curious about why/how this shockwaves occurred after the supersonic speed and, would like to know why the shock moved downstream as the speed was increased?
Thanks

At subsonic speeds, the air striking for example the nose cone of a plane can communicate the approach of the plane to the air ahead of it with a displacement wave that can move no faster than the local speed of sound. In this way, the incoming flow can start to split some distance ahead of the nose cone itself and get started moving out of the way before the cone gets there.
AT the speed of sound, the displacement wave cannot propagate forward and instead it piles up into a thin layer right at the point of the cone. Within that layer, pressure and density go through a discontinuous jump which is called a shock wave.
Now, note that the speed of air flowing over the surface of a plane changes at different points along the plane's profile, which means that while the flow might not be supersonic right at the tip of the nose cone, it might well be halfway along the chord length of the wing. Now note that as the plane's speed is increased, the air velocity distribution over the skin of the plane will shift around in response and this effect will cause the shock waves (which indicate where the local flow goes supersonic) to also move around in ways which are difficult to predict.

Hi Niels, great answer. For my 2nd question please (on the shock motion), could pressure gradient have something to do with this? As it would take longer to slow down the higher velocity to its shock?

the guy here who knows everything about this topic is Peter Kampf, perhaps he will weigh in?